Geared Motor - Torque requirements

I'm trying to choose a 12vdc geared motor for a TV lift system for an RV. It's mounted in an overhead drawer.

I'm an ME but I've never had to deal with motors or the physics involved. I've done a little research and reviewed the principles but I'm not sure about my results or the method I've used to get there.

Am I missing any dynamics on the profile of the lift? I'm seeing it as a pivoting diagonal lift? Are motors rated by a straight vertical lift? Because it's a hinged lift, will the lift angle make any considerable difference in my torque requirements?

Can anyone tell me if I've gone wrong or missed anything?

[U]Formula used: τ[Nm]=(mass[kg]× g) × pulley radius[m][/U]

TV mass = 4.5kg
g = 9.81
Pulley radius = (12mm dia / 2) = 6mm / 1000 = .006m

τ[Nm]= (4.5 x 9.81) x .006

τ[Nm]= [B].26487Nm[/B] (264.87mNm or [B]2.7kgf cm[/B])



[U]Motor I've chosen based on 2.7 kgf cm:[/U]
RPM = 40
Rated Torque = 5.6 kg.cm
Max Torque = 24 kg.cm
Ratio reduction = 150
No-Load Current (ma) = ≤60
Rated Current (A) = ≤0.6
Stall Current (A) = 1.3

Here's the drawing. Note: the pin point is attached to the drawer.

[ATTACH]2600[/ATTACH]

What you are dealing with is called overhung load. This pdf should explain it much better than I can.

[ATTACH]2627[/ATTACH]


Great drawing but I am not certain how the raising/lowering is taking place (winding a cable?). You may have to change the driver or driven pulley to get the speed right. If you want to have easy speed change look into DC drives. Cheap and easy.